Question

solve the following limit..​

Answer 1

Explanation :

$\bigstar \: \tt lim_{x \to 0} \bigg( \cfrac{x}{ \sqrt{9 - x + x {}^{2} } - 3 } \bigg) \\ \\ \\ \underline{ \tt Applying \: L-hospital \: rule,\: }\\ \\ \\ \longrightarrow \tt \: lim_{x \to 0} \bigg( \cfrac{ \cfrac{d}{dx}(x) }{ \cfrac{d}{dx}( \sqrt{9 - x + x {}^{2} } - 3)} \bigg) \\ \\ \\ \longrightarrow \tt \: lim_{x \to 0} \bigg( \cfrac{1}{ \cfrac{ - 1 + 2x}{2 \sqrt{ x{}^{2} - x + 9} } } \bigg) \\ \\ \\ \longrightarrow \tt \: lim_{x \to 0} \bigg( \cfrac{2 \sqrt{x {}^{2} - x + 9}}{ - 1 + 2x} \bigg) \\ \\ \\ \longrightarrow \tt \bigg( \cfrac{2 \sqrt{0 {}^{2} - 0 + 9} }{ - 1 + 2 \times 0} \bigg) \\ \\ \\ \longrightarrow \tt \bigg( \cfrac{2 \sqrt{9} }{ - 1} \bigg) \\ \\ \\ \longrightarrow \tt \bigg( \cfrac{2 \times 3}{ - 1} \bigg) \\ \\ \\ \longrightarrow \tt \bigg( \cfrac{6}{ - 1} \bigg) \\ \\ \\ \longrightarrow \: \red{\boxed{ \underline{ \tt - 6 }}} \: \bigstar$

Hence :

$\dag \boxed{\tt lim_{x \to 0} \bigg( \cfrac{x}{ \sqrt{9 - x + x {}^{2} } - 3 } \bigg) =-6}$

Answer 2

Answer:

-6

Step-by-step explanation:

By substituting the limits directly, we get (0/0) which is an indeterminate quantity. Therefore we have to solve it using another method.

We will use method of rationalisation.

$\implies \lim \limits_{x \to 0} \dfrac{x}{ \sqrt{9 - x + {x}^{2} } - 3 }$

${ \implies \lim \limits_{x \to 0} \dfrac{x}{ \sqrt{9 - x + {x}^{2} } - 3 } \times \dfrac{ \sqrt{9 - x + {x}^{2} } + 3 }{\sqrt{9 - x + {x}^{2} } + 3 } }$

${ \implies \lim \limits_{x \to 0}\dfrac{ x(\sqrt{9 - x + {x}^{2} } + 3) }{(\sqrt{9 - x + {x}^{2} })^{2} - {(3)}^{2} } }$

${ \implies \lim \limits_{x \to 0}\dfrac{ x(\sqrt{9 - x + {x}^{2} } + 3) }{9 - x + {x}^{2} - 9} }$

${ \implies \lim \limits_{x \to 0}\dfrac{ x(\sqrt{9 - x + {x}^{2} } + 3) }{ {x}^{2} - x}}$

${ \implies \lim \limits_{x \to 0}\dfrac{ x(\sqrt{9 - x + {x}^{2} } + 3) }{x(x - 1)}}$

${ \implies \lim \limits_{x \to 0}\dfrac{ \sqrt{9 - x + {x}^{2} } + 3 }{x - 1}}$

${ \implies \dfrac{ \sqrt{9 - 0+ {0}^{2} } + 3 }{0 - 1}}$

${ \implies \dfrac{ \sqrt{9} + 3 }{- 1}}$

${ \implies \dfrac{ 3 + 3 }{- 1}}$

${ \implies - 6}$